Two masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ are connected by a massless spring of spring constant $\mathrm{k}$ and unstreched length $\ell$. The masses are placed on a frictionless straight channel - which we consider our $x$-axis. They are initially at rest at $\mathrm{x}=0$ and $\mathrm{x}=\ell$, respectively. At $\mathrm{t}=0$, a velocity of $\mathrm{v}_{0}$ is suddenly imparted to the first particle. At a later time to, the centre of mass of the two masses is at-

• A

  $\mathrm{x}=\frac{\mathrm{m}_{2} \ell}{\mathrm{m}_{1}+\mathrm{m}_{2}}$
• B

  $\mathrm{x}=\frac{\mathrm{m}_{1} \ell}{\mathrm{m}_{1}+\mathrm{m}_{2}}+\frac{\mathrm{m}_{2} \mathrm{v}_{0} \mathrm{t}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$
• C

  $\mathrm{x}=\frac{\mathrm{m}_{2} \ell}{\mathrm{m}_{1}+\mathrm{m}_{2}}+\frac{\mathrm{m}_{2} \mathrm{v}_{0} \mathrm{t}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$
• Correct answer
  D

  $\mathrm{x}=\frac{\mathrm{m}_{2} \ell}{\mathrm{m}_{1}+\mathrm{m}_{2}}+\frac{\mathrm{m}_{1} \mathrm{v}_{0} \mathrm{t}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

Solution

$$ \begin{array}{l} \mathrm{v}_{\mathrm{COM}}=\frac{\mathrm{m}_{1} \mathrm{v}_{0}+0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}, \mathrm{x}_{\mathrm{iCOM}}=\frac{\mathrm{m}_{1}(0)+\mathrm{m}_{2}(\ell)}{\mathrm{m}_{1}+\mathrm{m}_{2}} \\ \text { also } \mathrm{x}_{\mathrm{COM}}=\mathrm{x}_{\mathrm{COM}}+\mathrm{v}_{\mathrm{Com}} \mathrm{t} \\ \therefore \mathrm{x}_{\mathrm{COM}}=\left(\frac{\mathrm{m}_{2} \ell}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)+\frac{\mathrm{m}_{1} \mathrm{v}_{0} \mathrm{t}}{\mathrm{m}_{1}+\mathrm{m}_{2}} \end{array} $$

Hence, the correct answer is option: 'D'.